Proof that the Hodge Conjecture is False.

[DV]

Proof of the Hodge Conjecture - Final Write-Up.
by Philip White.
October 22, 2021


An “easily understood summary” will follow at the end.

I. PROBLEM STATEMENT DEFINITIONS

- A claim that I need to verify: One subset of the set of all Hodge Classes is the set of all valid morphisms (i.e., continuous functions) from, given all finitely bounded manifolds M in P^2, M to the complex plane. I.e., we define the subset P^2_HC of HC as follows: (P^2_HC (subset of) HC) = { X | M is a manifold in P^2 and X is a morphism from M to C } .
- Fix an arbitrary algebraic variety V (an algebraic variety is a set of solutions to a system of polynomial equations, in this case in P^2). A class is a set with a restriction on membership defined by a wf. Given a fixed arbitrary algebraic variety V, an algebraic cycle is a rational linear combination of classes of algebraic subvarieties of V. (The points are treated as vectors, and the "sum" of the classes represents all possible sums of vectors from the individual classes.)
- An m-manifold is a topological space such that for each point in the main set X, the neighborhood of that point in the space is homeomorphic to m-dimensional Euclidean space.
- Given a topological space X and a point p in X, a neighborhood of p is a subset V or X such that an open set U is such that p is an element of U and U is a subset of V.


II. SWISS CHEESE MANIFOLDS AND KEY CORRESPONDENCE FUNCTION.

Consider P^2. Think of an infinite piece of Swiss cheese (or an infinite standardized test scantron sheet with answer bubbles to bubble in), where every integer point pair (e.g., (5,3) , (7,7) , (8,6) , etc.) is surrounded by a small empty circular area with no points. The Swiss cheese piece is infinite; it doesn't matter that it is a subset of P^2 and not of R^2. We will fill in the full empty holes associated with each point that is an ordered pair of integers in the Swiss cheese piece based on certain criteria. Note that every point in the manifold is indeed in neighborhoods that are homeomorphic to 2-D Euclidean space, as desired (the Swiss cheese holes are perfect circles of uniform size, with radius 0.4).

Now, consider a fixed arbitrary subset S of Z x Z. We modify the Swiss cheese manifold in P^2, filling in each empty circular hole associated with each ordered pair that is an element of S in the Swiss cheese manifold, with all previously omitted points in the empty circular holes included; this could be thought of as “bubbling in some answers into the infinite scantron”. Let F1 : PowerSet(Z x Z) --> PowerSet(P^2) be this correspondence function that maps each subset of Z x Z to its associated Swiss cheese manifold.

Define an arbitrary morphism M : P^2_HC --> C, and let MS be the set containing all such valid functions M. Let the key correspondence function F2 : PowerSet (Z x Z) --> MS map every element S of PowerSet(Z x Z) to the least element of a well-ordering of the subset MS2 of MS such that all elements of MS2 are functions that map elements of F1(S) to the complex plane, which must exist due to the axiom of choice. (Note, we could use any morphism that maps a particular S.C. manifold to the complex plane. Also note, at least one morphism always exists in each case.)

For clarity: Basically, F2 maps every possible way to fill in the Swiss cheese holes to a particular associated morphism, such that this morphism itself maps the filled-in Swiss cheese manifold based on this filling-in scheme to the complex plane.


III. VECTOR AXIOMS, AND VECTOR INFERENCE RULE DEFINITIONS.

Now we define “vector axioms” and “vector inference rules.”

Each "vector axiom" is a “vector wf” that serves as an axiom of a formal theory and that makes a claim about the presence of a vector that lies in a rectangular closed interval in P^2, e.g, "v1 = <x,y>, where x is in [0 - 0.2, 0 + 0.2] and y is in [2 - 0.2, 2 + 0.2]". The lower coordinate boundaries (a=0 and b=2, here) must be integer-valued. The vector will be asserted to be a single fixed vector that begins at the origin, (0,0), and has a tail in the rectangular interval. The "vector formal theory inference rules” will be Turing machines based on rational-valued vector artihmetic—there are infinitely many such rules, of three types: 1) simple vector addition, 2) multiplication of a vector by a scalar integer, and 3) division of a vector by a scalar integer—that reject or accept all inputs, and never fail to halt; the output of these inference rules, given one or two valid axioms/theorems, is always another vector wf, which is a valid theorem. Note that class restrictions can be coded into these TMs; i.e., these three types of inference rules can be modified to exclude certain vector wfs from being theorems. The key "vector wfs” will always be in a sense of the form "v_k = <x,y> where the x-coordinate of v_k is in [a-0.2,a+0.2] and the y-coordinate of v_k is in [b-0.2,b+0.2] ". We will define the predicate symbol R1(a,b) to represent this, and simply define a large set of propositions of the form "R1(a,b)”, with a and b set to be fixed constant elements of the domain set of integers, as axioms. All axioms in a "vector formal theory" will be of this form, and each axiom can be used in proofs repeatedly. Given a fixed arbitrary class of algebraic cycles A, we can construct an associated "vector formal theory" such that every point in A that is present in certain areas of P^2 can be represented as a vector that is constructible based on linear combinations of and class restriction rules on, vectors. The key fact about vector formal theories that we need to consider is that for a set of points T in a space such that all elements of T are not elements of the classes of algebraic cycles, any associated vector wf W is not a theorem if the set of all points described by W is a subset of T. In other words, if an entire "window of points" is not in the linear combination, then the proposition associated with that window of points cannot be a theorem. Also, if any point in the "window of points" is in the linear combination, then the associated proposition is a theorem.

(Note: Each Swiss cheese manifold hole has radius 0.4, and the distance from the hole center to the bottom left corner of any vector-axiom-associated square region is sqrt(0.08), which is less than 0.4 .)

Importantly, given a formal vector theory V1, we treat all theorems of this formal theory as axioms of a second theory V2, with specific always-halting Turing-machine-based inference rules that are fixed and unchanging regardless of the choice of V1. This formal theory V2 represents the linear combinations of V1-based classes of algebraic cycles. The full set of theorems of V2 represents the totality of what points can and cannot be contained in the linear combination of classes of algebraic cycles.

The final key fact that must be mentioned is that any Swiss cheese manifold description can be associated with one unique vector formal theory in this way. That is, there is a one-to-one correspondence between Swiss cheese manifolds and a subset of the set of all vector formal theories. As we shall see, the computability of all such vector formal theories will play an important role in the proof of the negation of the Hodge Conjecture.


IV. THE PROPOSITION Q.

Now we can consider the proposition, "For all Hodge Classes of the (Swiss cheese) type described above SC, there exists a formal vector theory (as described above) with a set of axioms and a (decidable) set of inference rules such that (at least) every point that is an ordered pair of integers in the Swiss cheese manifold can be accurately depicted to be 'in the Swiss cheese manifold or out of it' based on proofs of 'second-level' V2 theorems based on the 'first-level' V1 axioms and first-level inference rules." That is: Given an S.C. Hodge Class and any vector wf in an associated particular vector formal theory, the vector wf is true if and only if there exists a point in the relevant Hodge Class that is in the "window of points" described by the wf.

It is important to note that the Hodge Conjecture implies Q. That is, if rational linear combinations of classes of algebraic cycles really can be used to express Hodge Classes, then we really can use vector formal theories, as explained above, to describe Hodge Classes.


V. PROOF THAT THE HODGE CONJECTURE IS FALSE.

Conclusion:

Assume Q. Then we have that for all Swiss-cheese-manifold Hodge Classes SC, the language consisting of 'second-level vector theory propositions based on ordered pairs of integers derived from SC that are theorems' is decidable. All subsets of the set of all ordered pairs of integers are therefore decidable, since each language based on each Hodge Class SC as described just above can be derived from its associated Swiss-Cheese Hodge Class and all subsets of all ordered pairs of integers can be associated with a Swiss-Cheese Hodge Class algebraically. In other words, elements of the set of subsets of Z x Z can be mapped to elements of the set of all Swiss-Cheese Hodge Classes with a bijection, whose elements can in turn be mapped to elements of a subset of the set of all vector formal theories with a bijection, which can in turn be mapped to a subset of the set all computable languages with a bijection, which can in turn be mapped to a subset of the set all Turing machines with a bijection. This implies that the original set, the set of all subsets of Z x Z, is countable, which is false. This establishes that the Hodge Conjecture is false, since: Hodge Conjecture —> Q —> (PowerSet(Z x Z is countable) and NOT PowerSet(Z x Z is countable)).


VI. EASILY UNDERSTOOD SUMMARY

A simple way to express the idea behind this proof is: We have articulated a logic-based way to express what might be termed “descriptions of rational linear combinations of classes of algebraic cycles.” These “descriptions” deal with “presence within a tile” in projective 2-D space of one or more points from a fixed rational linear combination of classes of algebraic cycles. This technique establishes that, when restricting attention to a particular type of Hodge Class, the Hodge Conjecture implies that there can only be countably infinitely many such “descriptions,” since each such description is associated with a computable language of “vector theorems” and thus a Turing machine. This leads to a contradiction, because there are uncountably infinitely many Swiss cheese manifolds and also uncountably infinitely many associated Hodge Classes derived from these manifolds, and yet there are only countably infinitely many of these mathematical objects if the Hodge Conjecture is true. That is why the Hodge Conjecture is false.

[DV]

I gave myself 48 hours to publish this proof, and it is correct. There is a slight revision that I just noticed needs to be made while re-reading the proof: The "vector wfs" that are theorems might, in many cases, need to be "boolean wfs," i.e., a propositional formula comprised of atomic wfs, instead of just individual atomic wfs every time. This is due to the possibility that integer scalar multiplications and divisions might cause "boundary of tile issues," leading to the need for boolean connectives to fully express some theorems; however, the proof is still correct, since the final vector formal theory associated with any given Swiss cheese manifold would merely have multiple (and finite) possibilities for the final description (/tiling) of the Swiss cheese manifold, instead of just one possibility. That doesn't impact the "cardinality problem" that drives the proof, so the proof is still correct once this (fairly) minor is taken into account.

[DV]

The proof is correct as written; my intuition was very strong and correct on this problem, as it always is when I work on mathematical proofs these days. I do refuse to write up the article in highly precise "journal-ready format," and will continue to do so. If I am able to write a non-fiction book about math, I would be happy to include many other proofs of unsolved problems in this book.

As a clarification: When I use the word "tiling," I do not mean that the "tile" fully covers the S.c.m. hole in P^2. Instead, the tile is a small square that lies within the S.c.m. hole, and the full collection of "tiles" does not take up the whole set of points in the manifold.

Nevertheless, as a description of the manifolds, this idea works. That is because using a representative vector (in the algebraic construction) that must fall within the appropriate tile or fail to exist within the appropriate tile, and then applying the various inference rules, still yields a valid and relevant (and computable) "picture" of the Swiss cheese manifold, i.e., data about its description. Every vector theorem in a valid formal vector theory must be satisfied (i.e., at least one of the tiles indicated in the theorem must be "on"--there are no negations in these boolean propositions) to identify the appropriate "small tile" within each S.c.m. hole in the overall manifold.

This information is provided for those who might be looking at the proof to evaluate it and understand it.

[DV]

OK, there is one other minor mistake to fix to make this proof properly written: Basically, the closed rectangular (square) intervals need to be cut in half so that they are 0.1 * 2 = 0.2 in terms of the lengths of each side of the interval. The intervals are still closed, and actually, we do need every point in the Hodge Class to be represented, i.e., we do need a full tiling of the space. If there are overlaps on the boundary of two intervals, then if a vector touches this point, it must be counted as being in both of the overlapping intervals.

We still consider each closed interval that is centered around a point at (a,b) in Z x Z to be the one that represents that S.c.m. hole; in looking at the description of the S.c.m., we ignore the non-included intervals. The proof still works just fine...everything that needs to be computable is computable.

No I'm not motivated to write it up perfectly...why should I be? I thought it was perfectly right at first and found a few glitches; I chose not to "write out every step" in the conventional step, instead choosing to "write out every step" my own way since it's not being published. I will think about it a few more times and see if there are any other minor mistakes that I could fix...I think the proof is glitch-free at this point, but if it isn't and I find more issues, I might choose to write about them. (If I felt like it, yes, I could write it out in Theorem/Lemma + Proof format--writing each key theorem close to a well-formed formula and writing a clear and methodical argument for why it is correct, but that isn't going to happen for unpublishable work.)

Yes, admittedly, I thought the proof was 100% perfect the first and second times I wrote about it. The point at this point is that there are no unfixable errors, as before. The other point is, I am happy to claim since it's in my interests (and in the interests of other people who deserve better treatment): It is literally a "massively lethal error" that is the fault of the US government to illegally cause me to feel unmotivated. The CIA--the worst version of the worst organization in the US government history--is maximally to blame for this, and deserves bad consequences. That's right, I'm placing the blame on the CIA. It's not as though anyone else who has ever lived could handle either the CIA or proofs related to open problems in mathematics anywhere close to how well I handle them.

Indeed, if I "went all out" to work really hard and do the serious work to have "the best write-up ever," the CIA would see to it that I suffered for this effort. That is what happened when evil CIA pawn Eva Tardos got ahold of my relativization-related proof. I maintain that I am functioning optimally given the circumstances, in the sense that no one who has ever lived has performed as well at mathematics as I have done and as I may be continuing do, a little bit at least, if I feel like it.

I'm sorry that Americans who continue to support the CIA are so into organized seppuku. (I do mean it and I'm phrasing it a bit meanly. Please consider thinking about why I want to say it like that.)

[Malcolm McLean]

On Saturday, 23 October 2021 at 14:49:12 UTC+1, DV wrote:
> Proof of the Hodge Conjecture - Final Write-Up.
> by Philip White.
> October 22, 2021
>

> VI. EASILY UNDERSTOOD SUMMARY
>
> A simple way to express the idea behind this proof is: We have articulated a logic-based way to express what might be termed “descriptions of rational linear combinations of classes of algebraic cycles.” These “descriptions” deal with “presence within a tile” in projective 2-D space of one or more points from a fixed rational linear combination of classes of algebraic cycles. This technique establishes that, when restricting attention to a particular type of Hodge Class, the Hodge Conjecture implies that there can only be countably infinitely many such “descriptions,” since each such description is associated with a computable language of “vector theorems” and thus a Turing machine. This leads to a contradiction, because there are uncountably infinitely many Swiss cheese manifolds and also uncountably infinitely many associated Hodge Classes derived from these manifolds, and yet there are only countably infinitely many of these mathematical objects if the Hodge Conjecture is true. That is why the Hodge Conjecture is false.
>

I don't have the ability to tell you if this is correct, probably not a proof but on an interesting line, or a
complete misunderstanding. It's disappointing that no-one else seems to know either.

[DV]

Why do you assert that it is a "probably not a proof" of the stated proposition if you don't understand the math? It is definitely a proof. If you think math and science are important at all, why do you assert the contrary given the circumstances?

All anyone needs to know to understand this proof are these concepts:

- morphisms (continuous functions in topology, I've read) and basic topology (an understanding of morphisms, homeomorphisms, and what properties they have would be useful here)
- Turing machines and very basic TCS (you don't even need the fact that the halting problem is undecidable!)
- the definitions from the H.C. paper, which I tried to interpret and include
- basic mathematical reasoning, particularly relating to proofs of simply expressed theorems, and also a little bit of basic geometry/trigonometry
- an understanding of formal logic and how formal theories work (axioms, inference rules, proofs, etc.)
- an understanding of sets and functions, including what a power set is (it's the set of all subsets of a set)
- an understanding of the fact (not even the proof) that the real numbers and natural numbers are not in bijection (the power set of the naturals has the same cardinality as the set of all real numbers)
- searched-for definitions of some advanced abstract algebra terms like variety, subvariety, cycle...also, the idea of linear combinations is important
- a basic understanding of vectors (understanding vector spaces would be helpful)

Although that may sound like a long list, I argue that all of these concepts could be understood by a bright sophomore in high school who would put in the time to research these concepts on the internet (it might take 72 hours total?...working exercises would probably be optional to gain adequate comprehension?), except for maybe #4: basic mathematical reasoning. An understanding of #5 might be a little bit challenging, too. If you want to understand the proof, I recommend that you try to do some basic proofs as exercises (e.g., you could prove that the halting problem is unsolvable, that there are infinitely many primes, that the reals are not in bijection with the natural numbers, and that the square root of 2 is irrational...all of those proofs are fairly easy to do) and maybe read (and work exercises in?) Chapters 1 and 2 in Mendelson's "Introduction to Mathematical Logic."

[DV]

(I forgot to say "manifolds," also.)

[DV]

(By the way, although you don't need proofs by induction to understand this proof, you would need to understand that concept to be mathematically literate. If you read about on the internet and work some of the exercises in Mendelson (I think Chapter 1 has some good ones) that are about (non-fully formal) proofs by induction, you will probably gain competence at proofs by induction pretty quickly, since the exercises are "more challenging than average" I would argue (without certainty), compared to induction proofs that you might see in an undergraduate abstract algebra or discrete math class.)

(Also in case you are wondering, no, nothing I've written on the internet is likely to get anyone up to speed at math to the point where such a person would be competitive with me. I'm just discussing the basics; my math abilities are related to my main competitive secret or important skill or whatever.)